A 2 cm candle is placed in front of convex mirror having focal length of 15 cm . Image is formed 10 cm away from the mirror . Find object distance ?
a) 6 cm
b) -30 cm
c) 30 cm
d) 6 cm
With explanation
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A 2 cm candle is placed in front of convex mirror having focal length of 15 cm . Image is formed 10 cm away from the mirror . Find object distance ?
a) 6 cm
b) -30 cm
c) 30 cm
d) 6 cm
With explanation
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Verified answer
Answer :-
Distance of the object is -30 cm from the mirror . [Option.b]
Explanation :-
We have :-
→ Height of the object = 2 cm
→ Focal length of the mirror = 15 cm
→ Position of the image = 10 cm
________________________________
Since the mirror is convex, we have :-
• f = + 15 cm
• v = + 10 cm
Now, according to mirror formula :-
1/v + 1/u = 1/f
⇒ 1/u = 1/f - 1/v
⇒ 1/u = 1/15 - 1/10
⇒ 1/u = (1 - 2)/30
⇒ 1/u = -1/30
⇒ -u = 30
⇒ u = - 30 cm
Note : Since the object, by convention is always placed to the left of the mirror, it's x-coordinate 'u' is always negative.
Given : Height of candle is ( h ) 2 cm , Focal Length of the mirror is ( f ) 15 cm , Distance between mirror and image is ( v ) 10 cm & The mirror is Convex Mirror .
Exigency To Find : The object distance ( u ) .
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
⠀⠀⠀⠀⠀Given that ,
⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀The mirror is CONVEX MIRROR .
Therefore,
⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀ Focal Length of the mirror is ( f ) 15 cm ,
⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀ Distance between mirror and image is ( v ) 10 cm
[tex]\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\:\bf Formula\:of\:mirror\:: \\[/tex]
[tex]\qquad \dag\:\:\bigg\lgroup \sf{ \qquad \dfrac{1}{v}\:\: +\:\: \dfrac{1}{u}\:\:=\:\:\dfrac{1}{f}\qquad }\bigg\rgroup \\\\[/tex]
⠀⠀⠀⠀⠀Here , u is the object distance , v is the image distance & f is the focal length.
[tex]\qquad:\implies \sf \dfrac{1}{v} + \dfrac{1}{u}\:=\:\dfrac{1}{f} \\\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\[/tex]
[tex]\qquad:\implies \sf \dfrac{1}{v} + \dfrac{1}{u}\:=\:\dfrac{1}{f} \\\\[/tex]
[tex]\qquad:\implies \sf \dfrac{1}{10} + \dfrac{1}{u}\:=\:\dfrac{1}{15} \\\\[/tex]
[tex]\qquad:\implies \sf \dfrac{1}{u}\:=\:\dfrac{1}{15} - \dfrac{1}{10}\\\\[/tex]
[tex]\qquad:\implies \sf \dfrac{1}{u}\:=\: \dfrac{1- 2}{30}\\\\[/tex]
[tex]\qquad:\implies \sf \dfrac{1}{u}\:=\: \dfrac{- 1 }{30}\\\\[/tex]
[tex]\qquad:\implies \sf u\:=\: -30\\\\[/tex]
[tex]\qquad :\implies \pmb{\underline{\purple{\:u = -30\: cm }} }\:\:\bigstar \\\\[/tex]
⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀Here , u denotes object distance which is - 30 cm .
⠀⠀⠀⠀[tex]\therefore {\underline{ \sf \:Hence,\:The\:Object \:Distance \:(\ u \ ) \: \:is\:\bf - 30\:cm}}\\[/tex]